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What happens to the force between the charges if the distance between the charges is increased to two times its original value?

Answer 1

Julia Aguilar

#F_2 = 1/4 F_1#

The electrostatic force between charges follows Coulomb's Law:

#vec F = (k Q_1 Q_2)/r^2 hat r#

We see that for 2 charges, #F propto 1/r^2#

Or #F_2 = F _1 r_1^2/r_2^2 #

With #r_2 = 2 r_1# we have

#F_2 = F _1 r_1^2/(2r_1)^2 = 1/4 F_1#

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Answer 2

Nora Arzate

If the distance between charges is increased to two times its original value, the force between the charges decreases to one-fourth of its original value. This relationship is described by Coulomb's law, which states that the force between two charges is inversely proportional to the square of the distance between them. Therefore, doubling the distance results in the force being reduced by a factor of four.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.